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\author{五六七}
\title{常微分方程第2周作业（2.2-2.3）}

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\begin{document}

\maketitle

% \abstract{
% 两个实际例子：RL电路、追线问题。
% 2.2: 1(1), 2(1), 3(1)。 
% 2.3: 1(1), 2(1)。 
% }

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\begin{enumerate}

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\item RL电路：见DEBE目录。

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\item 追线问题：见DEBE目录。

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\item  (2.2)\#1(1): 
求解下列微分方程，并指出在平面上有意义的区域：
$\frac{dy}{dx} = \frac{x^2}{y}$. 

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item 分离变量可得 $ydy=x^2dx$. 
\item 两边积分可得 $\frac{1}{2}y^2=\frac{1}{3}x^3+C$. 
\item 原方程的定义区域是 $y\neq 0$. 
\end{enumerate}
}

\vspace{0.5cm}

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\item  (2.2)\#2(1): 
求解下列微分方程的初值问题：
$\sin 2x dx + \cos 3y dy = 0, \,\, y(\pi/2) = \pi/3$. 

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{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item 分离变量可得 $\sin 2x dx = - \cos 3y dy$. 
\item 两边积分可得 $-\frac{1}{2}\cos 2x = \frac{1}{3} \sin 3y +C$. 
\item 代入 $x=\pi/2, y=\pi/3$ 可得 $C=\frac{1}{2}$. 
\item 初值问题的解为 $-\frac{1}{2}\cos 2x = \frac{1}{3} \sin 3y +\frac{1}{2}$. 
\end{enumerate}

}

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\item  (2.2)\#3(1): 
求解下列微分方程，并作出相应的积分曲线族的简图：
$\frac{dy}{dx} = \cos x$. 

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{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item 分离变量可得 $dy = \cos x dx$. 
\item 两边积分可得 $y = \sin x +C$, 其中 $C$ 是任意常数。
\item 积分曲线是正弦函数 $y=\sin x$, 上下平移任意距离 $C$ 得到其它积分曲线。
\end{enumerate}

}

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\item  (2.3)\#1(1): 
求解微分方程：
$\frac{dy}{dx} + 2y = xe^{-x}$. 

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{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item 两边乘以积分因子 $e^{2x}$ 可得 $e^{2x}(\frac{dy}{dx} + 2y) = xe^{x}$. 
\item 左边化为 $(e^{2x}y)'=xe^x$. 
\item 两边积分可得 $e^{2x}y=\int xe^xdx$. 
\item 可得 $e^{2x}y=xe^x-e^x+C$, 可得 $y=xe^{-x}-e^{-x}+Ce^{-2x}$. 
\end{enumerate}

}

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\item  (2.3)\#2(1): 
把下列微分方程化为线性微分方程：
$\frac{dy}{dx} = \frac{x^2+y^2}{2y}$. 

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{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item 设 $u=y^2$, 则原方程化为 $\frac{du}{dx}=x^2+u$. 
\item 这是一阶线性常微分方程。
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\end{enumerate}

}

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\end{enumerate}

\end{document}

